I would like to take the derivative of this inverse function at $\pi$: $f(x) = 2x + \cos{x}$, given that ${f}^{-1}(\pi) = \frac{\pi}{2}$.

I know that there are two methods of doing it. Let me demonstrate the method that I have down pat, using the fact that $\frac{d}{dx}\left[{f}^{-1}(x)\right] = \frac{1}{{f}^{\prime}\left({f}^{-1}(x)\right)}$.

The problem is that method 2 is very inefficient. As you already notice, isolating $y$ from the equation $x=2y+cosy$ is really hard (it might be actually impossible in some cases).
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azarelJan 27 '12 at 4:35

Yes... but for my Calc II class I'm required to know the second method. Hope they won't ask to method 2 on a function like this. Yikes!!!
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Oliver SprynJan 27 '12 at 4:47

Some of the lines in your first solution are technically wrong, though you arrive at the right answer. In lines $4$ to the end, it looks as if you are differentiating a constant. The derivative of a constant is $0$. The simplest notational trick is to say let $g(x)=f^{-1}(x)$. Then $g'(x)=\dots$. Therefore $g'(\pi)=\dots$. The reason is that it is awkward to put a "prime" on $f^{-1}(x)$.
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André NicolasJan 27 '12 at 6:33

1 Answer
1

It is known that an inverse function $exists$ for any one-to-one function, but in many cases it cannot be expressed in terms of elementary functions. So, your first calculation may be the best you can do without using more machinery.